-3t^2+90t-432=0

Solution for -3t^2+90t-432=0 equation:

-3t^2+90t-432=0

a = -3; b = 90; c = -432;
Δ = b2-4ac
Δ = 902-4·(-3)·(-432)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-54}{2*-3}=\frac{-144}{-6}
=+24
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+54}{2*-3}=\frac{-36}{-6}
=+6
$